\beginsection{14.1}

Determine which of the following series converge.
Justify your answers.

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(a) $\sum n^4/2^n$
\medskip
Try the ratio test.
$$
{a_{n+1}\over a_n}
={(n+1)^4/2^{n+1}\over n^4/2^n}
={(n+1)^4\over2^{n+1}}\times{2^n\over n^4}
={(n+1)^4\over2n^4}\rightarrow{1\over2}<1
$$
So by the ratio test $\sum n^4/2^n$ converges.

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(b) $\sum2^n/n!$
\medskip
Try the ratio test.
$$
{a_{n+1}\over a_n}={2^{n+1}/(n+1)!\over 2^n/n!}
={2^{n+1}\over(n+1)!}\times{n!\over2^n}={2\over n+1}<1
$$
So by the ratio test $\sum2^n/n!$ converges.

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(c) $\sum n^2/3^n$
\medskip
Try the ratio test.
$$
{a_{n+1}\over a_n}
={(n+1)^2/3^{n+1}\over n^2/3^n}
={(n+1)^2\over 3^{n+1}}\times{3^n\over n^2}
={(n+1)^2\over3n^2}\rightarrow{1\over3}<1
$$
So by the ratio test $\sum n^2/3^n$ converges.

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(d) $\sum n!/(n^4+3)$
\medskip
Try the ratio test.
$$
{a_{n+1}\over a_n}
={(n+1)!/((n+1)^4+3)\over n!/(n^4+3)}
={(n+1)!\over((n+1)^4+3)}\times{n^4+3\over n!}
\rightarrow n+1>1
$$
So by the ratio test $\sum n!/(n^4+3)$ diverges.

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(e) $\sum\cos^2 n/n^2$
\medskip
Use the comparison test.
Since $\sum1/n^2$ converges and
$|\cos^2 n/n^2|\le1/n^2$ for all $n$, $\sum\cos^2 n/n^2$ must also converge.

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(f) $\sum_{n=2}^\infty1/(\log n)$
\medskip
Use the comparison test.
Since $\sum(1/n)$ diverges and $1/(\log n)>1/n$ for all $n>1$,
$\sum_{n=2}^\infty1/(\log n)$ must also diverge.